What is the integral of (x^3)/(sqrt(1-x^2)) ?

The integral of (x^3)/(sqrt(1-x^2)) is 1/3 (1-x^2)^{3/2}-sqrt(1-x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^3)/(sqrt(1-x^2))

\int \frac{x ^{3}}{1 - x ^{2}} d x

\frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2} + C

Solution steps

\int \frac{x ^{3}}{1 - x ^{2}} d x

Apply u-substitution

: \int u^{2} - 1 d u

= \int u^{2} - 1 d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int u^{2} d u - \int 1 d u

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

= \frac{u ^{3}}{3} - u

Substitute back

u = 1 - x^{2}

= \frac{( 1 - x ^{2} ) ^{3}}{3} - 1 - x^{2}

Simplify

\frac{( 1 - x ^{2} ) ^{3}}{3} - 1 - x^{2} : \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2}

= \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2}

Add a constant to the solution

= \frac{1}{3} (1 - x^{2})^{\frac{3}{2}} - 1 - x^{2} + C

\frac{e ^{- s}}{s ^{2}}

n = 0 \sum \infty e^{- 0.02 n}

(t + 7)^{3}

\frac{t - t}{t ^{\frac{1}{6}}}

(2 x - 1)^{2}

What is the integral of (x^3)/(sqrt(1-x^2)) ?
The integral of (x^3)/(sqrt(1-x^2)) is 1/3 (1-x^2)^{3/2}-sqrt(1-x^2)+C